Nguyên hàm

Ta có: \(\int{f\left( x \right)dx}=\int{\cos 3xdx}=\dfrac{\sin 3x}{3}+C\)

Ta có \(\int{f\left( x \right)\,\text{d}x}=\int{\left( \sin 5x+2 \right)\,\text{d}x}\) \(=-\dfrac{1}{5}\cos 5x+2x+C.\)

\(\int\limits_{{}}^{{}}{\cos 2xdx}=\dfrac{\sin 2x}{2}+C\)

\(\int{f(x)dx}=\int{2\cos 2x}dx\) \( = 2.\dfrac{{\sin 2x}}{2} + C = \sin 2x + C\)

Ta có \(\int{\sin 2x\,\text{d}x}\) \(=-\dfrac{\cos 2x}{2}+C.\)

Ta có: \(\int{\cos xdx=\sin x+C.}\)

Ta có \(f\left( x \right) = \dfrac{{3x - 2}}{{{{\left( {x - 2} \right)}^2}}} = \dfrac{{3x - 6 + 4}}{{{{\left( {x - 2} \right)}^2}}}\) \( = \dfrac{3}{{x - 2}} + \dfrac{4}{{{{\left( {x - 2} \right)}^2}}}\)

\(\begin{array}{l} \Rightarrow \int {f\left( x \right)dx}  = \int {\left[ {\dfrac{3}{{x - 2}} + \dfrac{4}{{{{\left( {x - 2} \right)}^2}}}} \right]dx} \\ = 3\int\limits_{}^{} {\dfrac{{dx}}{{x - 2}}}  + 4\int\limits_{}^{} {\dfrac{{dx}}{{{{\left( {x - 2} \right)}^2}}}} \\ = 3\ln \left| {x - 2} \right| - \dfrac{4}{{x - 2}} + C\\Do\,\,x \in \left( {2; + \infty } \right) \Rightarrow x - 2 > 0\\ \Rightarrow \int {f\left( x \right)dx}  = 3\ln \left( {x - 2} \right) - \dfrac{4}{{x - 2}} + C\end{array}\)

Ta có \(\int{f\left( x \right)\,\text{d}x}=\int{\dfrac{1}{x}\,\text{d}x}=\ln \left| x \right|+C.\)

Ta có \(\int {\left( {{x^2} + 2} \right)dx = \dfrac{{{x^3}}}{3} + 2x + C.} \)

Ta có: \(\int {f\left( x \right)dx}  = \int {\left( {{e^x} + x} \right)dx}  = {e^x} + \dfrac{1}{2}{x^2} + C\)

Đáp án A: \(F'\left( x \right) = \sin x = f\left( x \right) \Rightarrow \) Đáp án A đúng.

Đáp án B: hiển nhiên đúng.

Đáp án C: \(\int\limits_{}^{} {\dfrac{{u'\left( x \right)}}{{u\left( x \right)}}dx}  = \ln \left| {u\left( x \right)} \right| + C \Rightarrow \) Đáp án C sai.

Đáp án D: \(F'\left( x \right) = 2x = f\left( x \right) \Rightarrow \) Đáp án D đúng.

Ta có \(\int{\dfrac{1}{x}\,\text{d}x}=\ln \left| x \right|+C\ne \ln x+C.\)

Ta có: \(\int {\left( {{e^x} + 4} \right)dx = {e^x} + 4x + C.} \)

Áp dụng công thức \(\int{{{a}^{x}}}=\dfrac{{{a}^{x}}}{\ln a}+C\) ta được:

\(\int{{{3}^{x}}}=\dfrac{{{3}^{x}}}{\ln 3}+C\)

\(f\left( x \right) = {e^{1 - 4x}} \Rightarrow \int\limits_{}^{} {f\left( x \right)dx}  =- \dfrac{{ 1}}{4}{e^{1 - 4x}} + C\)

\(\int{\left( x+{{2}^{x}} \right)}dx=\dfrac{{{x}^{2}}}{2}+\dfrac{{{2}^{x}}}{\ln a}+C\)

Ta có \(\int{f\left( x \right)\,\text{d}x}=\int{{{2018}^{x}}\,\text{d}x}=\dfrac{{{2018}^{x}}}{\ln 2018}+C.\)

Ta có \(I=\int{\left( {{2}^{x}}+{{3}^{x}} \right)\,\text{d}x}=\int{{{2}^{x}}\,\text{d}x}+\int{{{3}^{x}}\,\text{d}x}=\dfrac{{{2}^{x}}}{\ln 2}+\dfrac{{{3}^{x}}}{\ln 3}+C.\)

Ta có \(f\left( x \right)={{25}^{x}}\Rightarrow \int{f\left( x \right)\,\text{d}x}=\int{{{25}^{x}}\,\text{d}x}=\dfrac{{{25}^{x}}}{\ln 25}+C=\dfrac{{{5}^{2x}}}{2\ln 5}+C.\)

\(\int {f(x)dx} = \int {{e^{2018x}}dx} = \dfrac{1}{{2018}}{e^{2018x}} + C\)